How to solve $\lim_\limits{x\to1} \frac{\sqrt{2x-1} -1}{x^2-1}$?

Question

If I substitute 1 to all the $x$ I get $\frac{0}{0}$. So I thought to factorize the expression. I can factorize the denominator $x^2-1$ and it becomes $(x+1)(x-1)$ but I don't know what to do with the numerator. Any suggestions?

Answer 1

multiply numerator and denominator by $$\sqrt{2x-1}+1$$ you will get $$\frac{2(x-1)}{(x-1)(x+1)(\sqrt{2x-1}+1)}$$

Answer 2

$$\begin{array}{rcl} \displaystyle \lim_{x \to 1} \frac {\sqrt{2x-1} - 1} {x^2-1} &=& \displaystyle \lim_{x \to 1} \frac {(\sqrt{2x-1} - 1) (\sqrt{2x-1} + 1)} {(x^2-1) (\sqrt{2x-1} + 1)} \\ &=& \displaystyle \lim_{x \to 1} \frac {(2x-1)-1} {(x-1) (x+1) (\sqrt{2x-1} + 1)} \\ &=& \displaystyle \lim_{x \to 1} \frac {2x-2} {(x-1) (x+1) (\sqrt{2x-1} + 1)} \\ &=& \displaystyle \lim_{x \to 1} \frac {2(x-1)} {(x-1) (x+1) (\sqrt{2x-1} + 1)} \\ &=& \displaystyle \lim_{x \to 1} \frac 2 {(x+1) (\sqrt{2x-1} + 1)} \\ &=& \displaystyle \frac 2 {(1+1) (\sqrt{2-1} + 1)} \\ &=& \displaystyle \frac 2 4 \\ &=& \displaystyle \frac 1 2 \\ \end{array}$$

Answer 3

If we multiply nominator and denominator by $\sqrt{2x-1}+1$, we get $$\lim_{x \to 1}\frac{2x-1-1}{(\sqrt{2x-1}+1)(x^2-1)} = \lim_{x \to 1}\frac{2(x-1)}{(\sqrt{2x-1}+1)(x+1)(x-1)} = \frac{1}{2}$$

Answer 4

$$\lim_{x\to1}\frac{\sqrt{2x-1}-1}{x^2-1}=\lim_{x\to1}\frac1{x+1}\cdot\frac{\sqrt{2x-1}-1}{x-1}=\frac{f'(1)}2,$$where $f(x)=\sqrt{2x-1}$.

Answer 5

Another possibility than multiplying by the conjugated quantity as in the other answers, is to study $f(1+u)$ when $u\to 0$ since $x\to 1\iff x=1+u$ with $u\to 0$.

$f(1+u)=\dfrac{\sqrt{2+2u-1}-1}{(1+u)^2-1}=\dfrac{\sqrt{1+2u}-1}{2u+u^2}=\dfrac{1+u+o(u)-1}{2u+o(u)}\to \dfrac 12$

Answer 6

Let $\sqrt{2x-1}-1=2y\implies x=2y^2+2y+1$